zoj 1151 Word Reversal(字符串操作模拟)

题目连接：

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1151

题目描述：

For each list of words, output a line with each word reversed without changing the order of the words.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

You will be given a number of test cases. The first line contains a positive integer indicating the number of cases to follow. Each case is given on a line containing a list of words separated by one space, and each word contains only uppercase and lowercase letters.

Output

For each test case, print the output on one line.

Sample Input

1

3
I am happy today
To be or not to be
I want to win the practice contest

Sample Output

I ma yppah yadot
oT eb ro ton ot eb
I tnaw ot niw eht ecitcarp tsetnoc

/*问题 将一段文字的每个单词反转，但次序不变
解题思路 模拟，具体算法见注释处的坑点*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cctype>
#include<string>
#include<algorithm>
using namespace std;

int main()
{
    int T,t;
    char str[101],str1[101];
    string s;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&t);
        getchar();
        while(t--){
            cin.getline(str,101);
            
            int len=strlen(str);
            int i;
            for(i=len-1;i>=0;i--)
                if(isalpha(str[i])) break;
            str[i+1]=' ';
            str[i+2]='';    
            
            char *p=NULL;
            p=str;
            
            while(*p == ' ')
                p++;
            
            len=strlen(p);
            for(i=0;i<len;i++){
                if(p[i]!=' ')
                {
                    s += p[i];
                }
                else
                {
                    reverse(s.begin(),s.end());
                    cout<<s;
                    if(i != len-1) printf(" ");//坑点1 
                    s="";
                }
            }
            cout<<endl;
            s="";
        }
        if(T != 0) printf("
");//坑点2 
    }
    return 0;
}